An original Thinka practice paper modelled on the structure and difficulty of the Nov 2025 (V4) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
卷一 (選擇題)
Answer all 40 multiple-choice questions. For each question, choose the one correct answer from the four options A, B, C or D.
40 題目 · 40 分
題目 1 · 選擇題
1 分
A solid metal cylinder has mass \( m = (25.0 \pm 0.1)\text{ g} \), diameter \( d = (12.00 \pm 0.05)\text{ mm} \) and length \( h = (50.0 \pm 0.2)\text{ mm} \). What is the percentage uncertainty in the calculated density of the metal?
A.0.8%
B.1.2%
C.1.6%
D.2.0%
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解題
The formula for density is \( \rho = \frac{m}{V} \) where volume \( V = \frac{\pi d^2 h}{4} \). Substituting this into the density formula gives \( \rho = \frac{4m}{\pi d^2 h} \). The fractional uncertainty in density is given by: \( \frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta h}{h} \). Calculating each term: \( \frac{\Delta m}{m} = \frac{0.1}{25.0} = 0.0040 \) (or 0.40%), \( 2 \frac{\Delta d}{d} = 2 \times \frac{0.05}{12.00} \approx 0.0083 \) (or 0.83%), and \( \frac{\Delta h}{h} = \frac{0.2}{50.0} = 0.0040 \) (or 0.40%). Adding these together: \( 0.40\% + 0.83\% + 0.40\% = 1.63\% \approx 1.6\% \).
評分準則
1 mark for calculating the percentage uncertainty of each individual term, correctly doubling the percentage uncertainty of the diameter, and summing them to obtain 1.6%.
題目 2 · 選擇題
1 分
An object is projected vertically upwards from a platform with an initial speed \( u \). It rises to its maximum height and then falls to the ground below, hitting it with speed \( v \). If the acceleration of free fall is \( g \) and air resistance is negligible, which expression gives the total time of flight \( T \) of the object?
A.\( \frac{v-u}{g} \)
B.\( \frac{v+u}{g} \)
C.\( \frac{\sqrt{v^2 - u^2}}{g} \)
D.\( \frac{\sqrt{v^2 + u^2}}{g} \)
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解題
Using the kinematic equation \( v = u + at \) with upwards as the positive direction: the initial velocity is \( +u \) and the final velocity when hitting the ground is \( -v \) (since it is moving downwards). The acceleration is \( -g \). Substituting these values: \( -v = u - gT \), which rearranges to \( gT = u + v \), and thus \( T = \frac{u+v}{g} \).
評分準則
1 mark for applying the equations of constant acceleration with correct signs for initial and final velocities to solve for the time of flight.
題目 3 · 選擇題
1 分
A heavy frame of weight \( W \) is suspended from a hook by a light symmetric string of total length \( L \) attached to two points at the top of the frame. The horizontal distance between these two attachment points is \( d \). What is the tension \( T \) in the string?
A.\( T = \frac{W d}{2\sqrt{L^2 - d^2}} \)
B.\( T = \frac{W L}{2\sqrt{L^2 - d^2}} \)
C.\( T = \frac{W L}{\sqrt{L^2 - d^2}} \)
D.\( T = \frac{W \sqrt{L^2 - d^2}}{2 L} \)
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解題
Each half of the string has length \( L/2 \) and supports a horizontal distance of \( d/2 \) from the center. This forms a right-angled triangle where the hypotenuse is \( L/2 \) and the horizontal base is \( d/2 \). The vertical height \( h \) of this triangle is \( h = \sqrt{(L/2)^2 - (d/2)^2} = \frac{1}{2}\sqrt{L^2 - d^2} \). Let \( \theta \) be the angle the string makes with the vertical. Then, \( \cos\theta = \frac{h}{L/2} = \frac{\sqrt{L^2 - d^2}}{L} \). For vertical equilibrium, the upward forces must balance the weight: \( 2 T \cos\theta = W \). Substituting for \( \cos\theta \) gives: \( 2 T \frac{\sqrt{L^2 - d^2}}{L} = W \), which rearranges to \( T = \frac{W L}{2\sqrt{L^2 - d^2}} \).
評分準則
1 mark for resolving the tension vertically, using geometry to express the angle in terms of L and d, and solving for tension T.
題目 4 · 選擇題
1 分
An electric motor is used to lift a load of mass \( 80\text{ kg} \) vertically through a height of \( 15\text{ m} \) at a constant speed in a time of \( 10\text{ s} \). The motor has an efficiency of \( 75\% \) and is connected to a \( 230\text{ V} \) power supply. What is the average current drawn by the motor?
A.3.8 A
B.5.1 A
C.6.8 A
D.9.1 A
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解題
The useful work output is \( W_{\text{out}} = mgh = 80 \times 9.81 \times 15 = 11772\text{ J} \). The useful power output is \( P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{11772}{10} = 1177.2\text{ W} \). Given the efficiency \( \eta = 0.75 \), the input electrical power is \( P_{\text{in}} = \frac{P_{\text{out}}}{0.75} = \frac{1177.2}{0.75} = 1569.6\text{ W} \). Since \( P_{\text{in}} = V I \), the current is \( I = \frac{1569.6}{230} \approx 6.82\text{ A} \), which rounds to \( 6.8\text{ A} \).
評分準則
1 mark for calculating useful power output, dividing by efficiency to find input power, and using the electric power equation to find the correct current.
題目 5 · 選擇題
1 分
Two wires, X and Y, are made of the same metal. Wire X has twice the diameter and half the length of wire Y. Both wires are suspended vertically and support the same load, which deforms them elastically. What is the ratio of the elastic potential energy stored in wire X to that in wire Y?
A.1:8
B.1:4
C.2:1
D.8:1
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解題
The elastic potential energy stored in a stretched wire is \( E_p = \frac{1}{2} F x \), where \( F \) is the load and \( x \) is the extension. Since both wires support the same load \( F \), the ratio of stored energy is equal to the ratio of extensions: \( \frac{E_{p,X}}{E_{p,Y}} = \frac{x_X}{x_Y} \). Extension is given by \( x = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E} \). Since both wires are made of the same metal, they have the same Young modulus \( E \). Therefore, \( x \propto \frac{L}{d^2} \). For wire X: \( L_X = 0.5 L_Y \) and \( d_X = 2 d_Y \). Thus, \( \frac{x_X}{x_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 = 0.5 \times \left(\frac{1}{2}\right)^2 = 0.125 = \frac{1}{8} \). The ratio is \( 1:8 \).
評分準則
1 mark for establishing that energy is proportional to extension for a constant force, expressing extension in terms of length and diameter squared, and calculating the correct ratio.
題目 6 · 選擇題
1 分
A stationary sound wave is established in a tube of length \( L \) that is closed at one end and open at the other. The wave is oscillating at its third harmonic (first overtone). Which of the following correctly describes the positions of the displacement nodes, measured from the closed end of the tube?
A.There are nodes at 0 and \( \frac{1}{2}L \).
B.There are nodes at 0 and \( \frac{2}{3}L \).
C.There are nodes at \( \frac{1}{3}L \) and \( L \).
D.There are nodes at \( \frac{1}{4}L \) and \( \frac{3}{4}L \).
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解題
For a tube of length \( L \) closed at one end, the closed end is always a displacement node, so there is a node at position \( x = 0 \). At the third harmonic, the length of the tube is equal to three-quarters of a wavelength, so \( L = \frac{3}{4}\lambda \), which means \( \lambda = \frac{4}{3}L \). Nodes occur at intervals of half a wavelength from the closed end. Therefore, the positions of the nodes are at \( x = 0 \) and \( x = \frac{\lambda}{2} = \frac{1}{2} \left(\frac{4}{3}L\right) = \frac{2}{3}L \).
評分準則
1 mark for identifying that the closed end is a displacement node and using the relationship between wavelength and tube length for the third harmonic to locate the second node.
題目 7 · 選擇題
1 分
A uniform wire of resistance \( R \) and length \( L \) is stretched uniformly until its length becomes \( 1.20 L \). During this process, the volume of the wire remains constant. What is the new resistance of the wire in terms of \( R \)?
A.\( 1.20 R \)
B.\( 1.38 R \)
C.\( 1.44 R \)
D.\( 1.56 R \)
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解題
Resistance is given by \( R = \rho \frac{L}{A} \). Since the volume \( V = A L \) remains constant, the cross-sectional area changes to \( A = \frac{V}{L} \). Substituting this into the resistance formula gives \( R = \rho \frac{L^2}{V} \). Since the resistivity \( \rho \) and volume \( V \) are constant, resistance is directly proportional to the square of the length: \( R \propto L^2 \). If the new length is \( 1.20 L \), then the new resistance is \( (1.20)^2 R = 1.44 R \).
評分準則
1 mark for relating resistance to the square of length when volume is constant, and correctly computing the squared factor.
題目 8 · 選擇題
1 分
During a beta-plus (\( \beta^+ \)) decay, a proton inside a nucleus decays into a neutron, a positron, and an electron neutrino. Which statement correctly describes the change in quark flavor inside the nucleon and the lepton number of the emitted neutrino?
A.An up quark changes to a down quark, and the emitted neutrino has a lepton number of \( +1 \).
B.An up quark changes to a down quark, and the emitted neutrino has a lepton number of \( -1 \).
C.A down quark changes to an up quark, and the emitted neutrino has a lepton number of \( +1 \).
D.A down quark changes to an up quark, and the emitted neutrino has a lepton number of \( -1 \).
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解題
A proton has a quark composition of \( uud \) and a neutron has a quark composition of \( udd \). During \( \beta^+ \) decay, one up quark (\( u \)) changes into a down quark (\( d \)). The emitted neutrino is an electron neutrino (\( \nu_e \)), which is a lepton (not an antilepton). Therefore, it has a lepton number of \( +1 \). This conserves lepton number because the other emitted lepton, the positron, is an antilepton with a lepton number of \( -1 \).
評分準則
1 mark for identifying the correct quark transformation (u to d) and the correct lepton number of the electron neutrino (+1).
題目 9 · 選擇題
1 分
To determine the density \(\rho\) of a uniform solid sphere, its mass \(m\) and radius \(r\) are measured. The measurements are as follows: \(m = (250 \pm 5)\text{ g}\) and \(r = (2.50 \pm 0.05)\text{ cm}\). What is the percentage uncertainty in the calculated density?
A.4.0%
B.5.0%
C.8.0%
D.14.0%
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解題
The density \(\rho\) of a sphere is given by \(\rho = \frac{m}{V} = \frac{3m}{4\pi r^3}\). The percentage uncertainty in density is calculated by adding the percentage uncertainty in mass to three times the percentage uncertainty in radius: \(\%\Delta\rho = \%\Delta m + 3\%\Delta r\). First, calculate the percentage uncertainties: \(\%\Delta m = \frac{5}{250} \times 100 = 2.0\%\) and \(\%\Delta r = \frac{0.05}{2.50} \times 100 = 2.0\%\). Therefore, the percentage uncertainty in density is \(2.0\% + 3(2.0\%) = 8.0\%\).
評分準則
1 mark for identifying the correct relationship of combining uncertainties for a quotient with a power of 3 and calculating the correct final value of 8.0%.
題目 10 · 選擇題
1 分
A uniform rigid rod of weight \(W\) and length \(L\) is suspended horizontally in equilibrium by two vertical strings attached at its ends. A block of weight \(2W\) is placed on the rod at a distance of \(\frac{1}{4}L\) from the left end. What is the ratio \(\frac{T_1}{T_2}\) of the tension in the left-hand string (\(T_1\)) to the tension in the right-hand string (\(T_2\))?
A.1.5
B.2.0
C.2.5
D.3.0
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解題
Take moments about the right end of the rod: \(T_1 \times L = (2W) \times \frac{3}{4}L + W \times \frac{1}{2}L\). Dividing by \(L\) gives \(T_1 = 1.5W + 0.5W = 2.0W\). Since the system is in translational equilibrium, the total upward force must equal the total downward force: \(T_1 + T_2 = 2W + W = 3W\). Substituting \(T_1 = 2W\) gives \(T_2 = 3W - 2W = W\). Thus, the ratio of the tensions is \(\frac{T_1}{T_2} = \frac{2.0W}{W} = 2.0\).
評分準則
1 mark for using the principle of moments to calculate both tensions and finding their correct ratio of 2.0.
題目 11 · 選擇題
1 分
An electric motor is used to raise a crate of mass \(120\text{ kg}\) vertically upwards at a constant speed of \(1.5\text{ m s}^{-1}\). The efficiency of the motor is \(75\%\). What is the electrical power input to the motor?
A.1.3 kW
B.1.8 kW
C.2.4 kW
D.3.1 kW
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解題
The useful power output \(P_{\text{out}}\) required to lift the crate at constant speed is \(P_{\text{out}} = F v = m g v = 120\text{ kg} \times 9.81\text{ m s}^{-2} \times 1.5\text{ m s}^{-1} = 1765.8\text{ W}\). Since efficiency is defined as \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}}\), the electrical power input \(P_{\text{in}}\) is \(P_{\text{in}} = \frac{1765.8\text{ W}}{0.75} = 2354.4\text{ W} \approx 2.4\text{ kW}\).
評分準則
1 mark for calculating the useful output power using mgv and dividing by the efficiency of 0.75 to get 2.4 kW.
題目 12 · 選擇題
1 分
In a double-slit interference experiment using light of wavelength \(600\text{ nm}\), bright fringes are observed on a screen with a fringe separation of \(2.4\text{ mm}\). The wavelength of the light is then changed to \(450\text{ nm}\), and the distance between the slits is halved. The distance from the slits to the screen remains unchanged. What is the new fringe separation?
A.1.6 mm
B.1.8 mm
C.3.2 mm
D.3.6 mm
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解題
Fringe separation is given by \(x = \frac{\lambda D}{a}\). Let the initial values be \(\lambda_1 = 600\text{ nm}\), \(a_1\), and \(x_1 = 2.4\text{ mm}\). The new values are \(\lambda_2 = 450\text{ nm} = 0.75\lambda_1\), and \(a_2 = 0.5a_1\). Thus, the new fringe separation is \(x_2 = \frac{\lambda_2 D}{a_2} = \frac{0.75\lambda_1 D}{0.5a_1} = 1.5 \times \frac{\lambda_1 D}{a_1} = 1.5 x_1 = 1.5 \times 2.4\text{ mm} = 3.6\text{ mm}\).
評分準則
1 mark for setting up the proportional relationship for the double-slit formula and correctly solving for the new fringe separation of 3.6 mm.
題目 13 · 選擇題
1 分
Two blocks, X and Y, are on a frictionless horizontal surface. Block X has mass \(m\) and is moving to the right with speed \(u\). Block Y has mass \(3m\) and is moving to the left with speed \(2u\). The blocks collide head-on and the collision is perfectly elastic. What are the velocities of the blocks after the collision?
A.X has speed 2.0u to the left, Y has speed 1.0u to the right
B.X has speed 2.5u to the left, Y has speed 0.5u to the right
C.X has speed 3.5u to the left, Y has speed 0.5u to the left
D.X has speed 3.5u to the right, Y has speed 1.5u to the left
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解題
Let the direction to the right be positive. The initial velocities are \(u_X = u\) and \(u_Y = -2u\). By conservation of momentum: \(m(u) + 3m(-2u) = m v_X + 3m v_Y \implies v_X + 3v_Y = -5u\). For a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: \(u_X - u_Y = v_Y - v_X \implies u - (-2u) = v_Y - v_X \implies v_Y - v_X = 3u\). Solving these simultaneous equations, we find \(v_Y = -0.5u\) (moving to the left with speed \(0.5u\)) and \(v_X = -3.5u\) (moving to the left with speed \(3.5u\)).
評分準則
1 mark for correctly applying conservation of linear momentum and the relative speed condition for a perfectly elastic collision to obtain the velocities.
題目 14 · 選擇題
1 分
Two wires, P and Q, are made of the same material. Wire P has length \(L\) and diameter \(d\). Wire Q has length \(2L\) and diameter \(2d\). Both wires are suspended vertically and carry equal loads. What is the ratio \(\frac{\text{strain of wire P}}{\text{strain of wire Q}}\)?
A.0.5
B.1.0
C.2.0
D.4.0
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解題
Since both wires are made of the same material, they have the same Young modulus \(E\). Young modulus is defined as \(E = \frac{\text{stress}}{\text{strain}}\), so \(\text{strain} = \frac{\text{stress}}{E} = \frac{F}{A E}\). The cross-sectional area of wire P is \(A_P = \frac{\pi d^2}{4}\) and the cross-sectional area of wire Q is \(A_Q = \frac{\pi (2d)^2}{4} = \pi d^2 = 4 A_P\). Since the tensile force \(F\) and the Young modulus \(E\) are the same for both wires, the strain is inversely proportional to the cross-sectional area. Therefore, the ratio of the strains is \(\frac{\text{strain}_P}{\text{strain}_Q} = \frac{A_Q}{A_P} = 4.0\).
評分準則
1 mark for recognizing that strain depends on stress and Young modulus, and correctly calculating the ratio of the areas to find the strain ratio of 4.0.
題目 15 · 選擇題
1 分
A tube of length \(42.5\text{ cm}\) is closed at one end. A stationary sound wave is set up inside the tube by a loudspeaker at the open end. When the frequency of the sound is gradually increased, a resonance is heard at a certain frequency. The next higher frequency at which resonance occurs is \(600\text{ Hz}\). Assuming the speed of sound in air is \(340\text{ m s}^{-1}\), what is the fundamental resonant frequency of the tube?
A.150 Hz
B.200 Hz
C.300 Hz
D.400 Hz
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解題
For a tube closed at one end, the resonant frequencies are odd harmonics of the fundamental frequency \(f_1\), given by \(f_n = n f_1\) where \(n = 1, 3, 5, \dots\). The fundamental frequency is \(f_1 = \frac{v}{4L} = \frac{340\text{ m s}^{-1}}{4 \times 0.425\text{ m}} = 200\text{ Hz}\). The next resonant harmonic is the third harmonic, \(f_3 = 3 f_1 = 3 \times 200\text{ Hz} = 600\text{ Hz}\). This matches the given higher resonance frequency. Therefore, the fundamental resonant frequency is indeed \(200\text{ Hz}\).
評分準則
1 mark for using the closed-pipe resonance relation to calculate the fundamental frequency of 200 Hz.
題目 16 · 選擇題
1 分
A \(\Xi^0\) baryon consists of one up (\(\text{u}\)) quark and two strange (\(\text{s}\)) quarks, while a \(\pi^+\) pion consists of an up (\(\text{u}\)) quark and an anti-down (\(\bar{\text{d}}\)) antiquark. What are the total charges of the \(\Xi^0\) baryon and the \(\pi^+\) pion in terms of the elementary charge \(e\)?
A.Charge of \(\Xi^0\) is 0; Charge of \(\pi^+\) is \(+1e\)
B.Charge of \(\Xi^0\) is 0; Charge of \(\pi^+\) is \(-1e\)
C.Charge of \(\Xi^0\) is \(-1e\); Charge of \(\pi^+\) is \(+1e\)
D.Charge of \(\Xi^0\) is \(+1e\); Charge of \(\pi^+\) is 0
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解題
The charge of an up quark is \(+\frac{2}{3}e\), a strange quark is \(-\frac{1}{3}e\), and a down quark is \(-\frac{1}{3}e\) (which makes the charge of an anti-down antiquark \(+\frac{1}{3}e\)). For the \(\Xi^0\) baryon (\(\text{uss}\)), the total charge is \(+\frac{2}{3}e + 2(-\frac{1}{3}e) = 0\). For the \(\pi^+\) pion (\(\text{u}\bar{\text{d}}\)), the total charge is \(+\frac{2}{3}e + (+\frac{1}{3}e) = +1e\).
評分準則
1 mark for correctly determining the quark charges and summing them to get a net charge of 0 for the baryon and +1e for the pion.
題目 17 · 選擇題
1 分
A student determines the density \(\rho\) of a uniform cylinder by measuring its mass \(m\), length \(L\), and diameter \(d\).
The measurements and their absolute uncertainties are:
Mass \(m = (125.0 \pm 0.5)\text{ g}\)
Length \(L = (8.00 \pm 0.08)\text{ cm}\)
Diameter \(d = (2.50 \pm 0.05)\text{ cm}\)
What is the percentage uncertainty in the calculated value of the density?
Now, sum the percentage uncertainties, remembering that the uncertainty in the diameter must be multiplied by 2 because it is raised to the second power:
1 mark for the correct calculation of individual percentage uncertainties and summing them with the factor of 2 for diameter to obtain \(5.4\%\).
題目 18 · 選擇題
1 分
A uniform sphere of weight \(12\text{ N}\) and radius \(15\text{ cm}\) is suspended from a smooth vertical wall by a light string of length \(10\text{ cm}\) attached to its surface. The line of action of the tension in the string passes through the center of the sphere.
What is the tension in the string?
A.\(7.2\text{ N}\)
B.\(9.6\text{ N}\)
C.\(15\text{ N}\)
D.\(20\text{ N}\)
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解題
Let \(T\) be the tension in the string, \(W\) be the weight of the sphere, and \(R\) be the horizontal normal reaction from the wall.
The geometry of the system forms a right-angled triangle: - The hypotenuse is the distance from the attachment point on the wall to the center of the sphere: \(L + r = 10\text{ cm} + 15\text{ cm} = 25\text{ cm}\). - The horizontal side is the radius of the sphere, \(r = 15\text{ cm}\). - The vertical side along the wall is \(\sqrt{25^2 - 15^2} = 20\text{ cm}\).
Let \(\theta\) be the angle between the string and the vertical wall. From the right-angled triangle: \(\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{20}{25} = 0.80\)
For vertical equilibrium of the sphere: \(T \cos\theta = W\) \(T \times 0.80 = 12\text{ N}\) \(T = \frac{12}{0.80} = 15\text{ N}\)
評分準則
1 mark for calculating the correct angle or ratio of the triangle and using vertical equilibrium to find the tension of \(15\text{ N}\).
題目 19 · 選擇題
1 分
An electric pump is used to raise water from a well that is \(15\text{ m}\) deep and discharge it through a pipe of cross-sectional area \(2.0 \times 10^{-3}\text{ m}^2\) at a speed of \(4.0\text{ m s}^{-1}\). The density of water is \(1000\text{ kg m}^{-3}\).
What is the minimum useful electrical power input to the pump if its efficiency is \(65\%\)?
A.\(0.81\text{ kW}\)
B.\(1.2\text{ kW}\)
C.\(1.8\text{ kW}\)
D.\(1.9\text{ kW}\)
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解題
First, calculate the mass of water discharged per second, \(\frac{\Delta m}{\Delta t}\):
\(\frac{\Delta m}{\Delta t} = \rho A v = 1000\text{ kg m}^{-3} \times (2.0 \times 10^{-3}\text{ m}^2) \times 4.0\text{ m s}^{-1} = 8.0\text{ kg s}^{-1}\)
The useful power output \(P_{\text{out}}\) consists of raising the water (potential energy) and giving it kinetic energy:
1 mark for finding both potential and kinetic energy terms, combining them, and dividing by the efficiency to get \(1.9\text{ kW}\).
題目 20 · 選擇題
1 分
In a Young’s double-slit experiment, light of wavelength \(\lambda\) is incident on two slits separated by a distance \(d\). Interference fringes of width \(x\) are observed on a screen at a distance \(D\) from the slits.
The experiment is now modified: the wavelength of the light is increased to \(1.2\lambda\), the slit separation is reduced to \(0.8d\), and the distance to the screen is doubled.
What is the new fringe width?
A.\(0.48x\)
B.\(0.75x\)
C.\(1.5x\)
D.\(3.0x\)
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解題
The fringe width \(x\) is given by the formula:
\(x = \frac{\lambda D}{d}\)
With the modifications, the new fringe width \(x'\) is:
1 mark for correctly applying the double-slit formula with the scaling factors to find the correct ratio of \(3.0\).
題目 21 · 選擇題
1 分
A trolley of mass \(2m\) travels at speed \(3v\) along a frictionless horizontal track. It collides with a stationary trolley of mass \(m\). After the collision, the two trolleys stick together and move with a common speed.
What fraction of the initial kinetic energy is lost as a result of the collision?
A.\(\frac{1}{9}\)
B.\(\frac{1}{3}\)
C.\(\frac{2}{3}\)
D.\(\frac{8}{9}\)
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解題
Using the principle of conservation of linear momentum:
\(p_i = p_f\)
\((2m)(3v) + 0 = (2m + m)v_f\)
\(6mv = 3mv_f \implies v_f = 2v\)
Now, find the initial and final kinetic energies: - Initial kinetic energy: \(E_{k,i} = \frac{1}{2}(2m)(3v)^2 = 9mv^2\) - Final kinetic energy: \(E_{k,f} = \frac{1}{2}(3m)(2v)^2 = 6mv^2\)
1 mark for using momentum conservation to find final speed, computing initial and final kinetic energies, and calculating the lost fraction as \(\frac{1}{3}\).
題目 22 · 選擇題
1 分
A stationary sound wave is set up in a tube of length \(L\) closed at one end. The frequency of the fundamental note (first harmonic) is \(f\).
The closed end of the tube is now opened so that the tube is open at both ends.
What is the frequency of the first overtone (second harmonic) of the open tube?
A.\(f\)
B.\(2f\)
C.\(3f\)
D.\(4f\)
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解題
For a tube of length \(L\) closed at one end, the fundamental wavelength \(\lambda_1\) is:
\(\lambda_1 = 4L\)
Therefore, the fundamental frequency \(f\) is:
\(f = \frac{v}{4L}\)
For a tube open at both ends, the wavelength of the \(n\)-th harmonic is:
\(\lambda_n = \frac{2L}{n}\)
The first overtone of an open tube is its second harmonic (\(n=2\)), so its wavelength \(\lambda_2\) is:
\(\lambda_2 = \frac{2L}{2} = L\)
Its frequency \(f_2\) is:
\(f_2 = \frac{v}{\lambda_2} = \frac{v}{L}\)
Comparing this to \(f\):
\(f_2 = 4 \left(\frac{v}{4L}\right) = 4f\)
評分準則
1 mark for finding the frequency of the first overtone of the open tube in terms of the fundamental frequency of the closed tube, resulting in \(4f\).
題目 23 · 選擇題
1 分
The \(\Sigma^-\right.\) baryon is a strange particle with charge \(-1e\). It decays into a neutron and a \(\pi^-\) meson.
What are the quark compositions of the \(\Sigma^-\) baryon and the \(\pi^-\) meson?
(The charge of the up quark is \(+\frac{2}{3}e\), the down quark is \(-\frac{1}{3}e\), and the strange quark is \(-\frac{1}{3}e\).)
A baryon is composed of three quarks. Since the \(\Sigma^-\) baryon has a charge of \(-1e\) and contains a strange quark, let's check the combinations: - \(\text{dds}\): Charge \(= (-\frac{1}{3}e) + (-\frac{1}{3}e) + (-\frac{1}{3}e) = -1e\). This is correct. - \(\text{uss}\): Charge \(= (+\frac{2}{3}e) + (-\frac{1}{3}e) + (-\frac{1}{3}e) = 0\). This is neutral.
A meson consists of a quark-antiquark pair. The \(\pi^-\) meson has a charge of \(-1e\). Let's check the combinations: - \(\text{d}\bar{\text{u}}\): Charge \(= (-\frac{1}{3}e) + (-\frac{2}{3}e) = -1e\). This is correct. - \(\text{u}\bar{\text{d}}\): Charge \(= (+\frac{2}{3}e) + (+\frac{1}{3}e) = +1e\). This is the \(\pi^+\) meson.
Therefore, the correct combination is \(\Sigma^- = \text{dds}\) and \(\pi^- = \text{d}\bar{\text{u}}\).
評分準則
1 mark for correctly identifying the quark compositions based on charge conservation and definitions of baryons/mesons.
題目 24 · 選擇題
1 分
Three identical resistors, each of resistance \(R\), are connected to a cell of electromotive force (e.m.f.) \(E\) and negligible internal resistance. Two of the resistors are connected in parallel with each other. This parallel combination is connected in series with the third resistor \(X\) and the cell.
What is the ratio \(\frac{\text{Power dissipated in } X}{\text{Total power dissipated in the circuit}}\)?
A.\(\frac{1}{4}\)
B.\(\frac{1}{3}\)
C.\(\frac{1}{2}\)
D.\(\frac{2}{3}\)
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解題
Let \(R\) be the resistance of each resistor.
The resistance of the parallel pair of resistors is:
1 mark for finding the total resistance of the circuit and applying power formulas to obtain the correct ratio of \(\frac{2}{3}\).
題目 25 · 選擇題
1 分
An experimenter measures the period \(T\) of a simple pendulum and the length \(L\) of the pendulum to determine the acceleration of free fall \(g\). The relation used is \(g = \frac{4\pi^2 L}{T^2}\). The percentage uncertainty in the measurement of \(L\) is \(1.5\%\), and the percentage uncertainty in the measurement of \(T\) is \(2.0\%\). What is the percentage uncertainty in the calculated value of \(g\)?
A.3.5%
B.5.5%
C.7.5%
D.9.5%
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解題
The relationship is given by \(g = \frac{4\pi^2 L}{T^2}\). For a product or quotient, the percentage uncertainties are added, with each being multiplied by the power to which the variable is raised. Therefore, the percentage uncertainty in \(g\) is given by \(\frac{\Delta g}{g} \times 100\% = \frac{\Delta L}{L} \times 100\% + 2 \times \left(\frac{\Delta T}{T} \times 100\%\right)\). Substituting the given values: \(\text{Percentage uncertainty in } g = 1.5\% + 2 \times 2.0\% = 5.5\%\).
評分準則
1 mark for adding the percentage uncertainty of length to twice the percentage uncertainty of period to obtain 5.5%.
題目 26 · 選擇題
1 分
A uniform picture frame of mass \(2.0\text{ kg}\) is suspended in equilibrium from a fixed point by two identical light strings. Each string makes an angle of \(35^\circ\) to the horizontal. What is the tension in each string?
A.10.0\text{ N}
B.12.0\text{ N}
C.17.1\text{ N}
D.34.2\text{ N}
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解題
For vertical equilibrium, the sum of the vertical components of the tension in both strings must equal the weight of the frame. The weight of the frame is \(W = mg = 2.0 \times 9.81 = 19.62\text{ N}\). The vertical component of the tension in each string is \(T \sin(35^\circ)\). Since there are two strings, \(2 T \sin(35^\circ) = 19.62\text{ N}\). Solving for the tension: \(T = \frac{19.62}{2 \sin(35^\circ)} = 17.1\text{ N}\).
評分準則
1 mark for equating the sum of the vertical components of tension to the weight of the frame and solving for tension to obtain 17.1 N.
題目 27 · 選擇題
1 分
An electric motor is used to lift a load of \(50\text{ kg}\) vertically upwards through a height of \(12\text{ m}\) in a time of \(8.0\text{ s}\). The efficiency of the motor is \(75\%\). What is the input electrical power to the motor?
A.550\text{ W}
B.740\text{ W}
C.980\text{ W}
D.1300\text{ W}
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解題
First, calculate the useful work output: \(W_{\text{out}} = mgh = 50 \times 9.81 \times 12 = 5886\text{ J}\). Next, calculate the useful power output: \(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{5886}{8.0} = 736\text{ W}\). Since the efficiency is 75%, the input electrical power is \(P_{\text{in}} = \frac{P_{\text{out}}}{0.75} = \frac{736}{0.75} = 981\text{ W}\), which rounds to \(980\text{ W}\) to two significant figures.
評分準則
1 mark for calculating the useful output power and dividing it by the efficiency to find 980 W.
題目 28 · 選擇題
1 分
A stone is projected vertically upwards from the ground with an initial speed of \(15\text{ m s}^{-1}\). Air resistance is negligible. What is the total time for which the stone is at a height of \(5.0\text{ m}\) or more above the ground?
A.0.38\text{ s}
B.1.1\text{ s}
C.2.3\text{ s}
D.3.1\text{ s}
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解題
Using the equations of motion under constant acceleration, \(s = ut + \frac{1}{2}at^2\) where \(u = 15\text{ m s}^{-1}\), \(a = -9.81\text{ m s}^{-2}\), and \(s = 5.0\text{ m}\). This gives the quadratic equation: \(5.0 = 15t - 4.905t^2\), which simplifies to \(4.905t^2 - 15t + 5.0 = 0\). Solving this quadratic equation yields two times: \(t_1 = 0.38\text{ s}\) (the time when the stone passes \(5.0\text{ m}\) going upwards) and \(t_2 = 2.68\text{ s}\) (the time when it passes \(5.0\text{ m}\) on its way down). The total time the stone is above \(5.0\text{ m}\) is \(t_2 - t_1 = 2.68 - 0.38 = 2.3\text{ s}\).
評分準則
1 mark for establishing the quadratic equation for time at 5.0 m and calculating the difference between the two solutions to obtain 2.3 s.
題目 29 · 選擇題
1 分
A wire of length \(L\) and cross-sectional area \(A\) is made of a material of Young modulus \(E\). When a load \(F\) is applied, the wire extends by \(x\). A second wire made of the same material has length \(2L\) and cross-sectional area \(3A\). If a load of \(2F\) is applied to this second wire, what is its extension?
A.\(\frac{3}{4} x\)
B.\(x\)
C.\(\frac{4}{3} x\)
D.\(3 x\)
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解題
The Young modulus is given by \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{F L}{A x}\). Rearranging this for extension gives \(x = \frac{F L}{A E}\). For the second wire: \(x_2 = \frac{F_2 L_2}{A_2 E}\). Since it is made of the same material, the Young modulus \(E\) remains the same. Substituting the new values: \(x_2 = \frac{(2F)(2L)}{(3A)E} = \frac{4}{3} \frac{F L}{A E} = \frac{4}{3}x\).
評分準則
1 mark for writing the expression for extension and correctly substituting the given ratios to show the new extension is 4/3 of the original.
題目 30 · 選擇題
1 分
Two coherent waves meet at a point in phase. The amplitude of one wave is \(2 A_0\) and the amplitude of the other wave is \(3 A_0\). The intensity of the wave with amplitude \(2 A_0\) is \(I_0\). What is the resultant intensity at the point?
A.\(2.25 I_0\)
B.\(3.25 I_0\)
C.\(6.25 I_0\)
D.\(13.0 I_0\)
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解題
Since intensity is proportional to the square of the amplitude (\(I \propto A^2\)), we can write \(I = k A^2\). For the wave with amplitude \(2 A_0\), we have \(I_0 = k (2 A_0)^2 = 4 k A_0^2\), which means \(k A_0^2 = \frac{I_0}{4}\). Since the two coherent waves meet in phase, their amplitudes add constructively to give a resultant amplitude of \(A_{\text{resultant}} = 2 A_0 + 3 A_0 = 5 A_0\). The resultant intensity is then \(I_{\text{resultant}} = k (5 A_0)^2 = 25 k A_0^2 = 25 \left(\frac{I_0}{4}\right) = 6.25 I_0\).
評分準則
1 mark for determining the resultant amplitude as 5 A0 and using the intensity-amplitude squared relationship to obtain 6.25 I0.
題目 31 · 選擇題
1 分
A radioactive source contains a nuclide that decays by emitting \(\beta^-\)\ particles. The initial activity of the source is \(3.2 \times 10^5\text{ Bq}\). After a time interval of \(6.0\text{ hours}\), the activity has decreased to \(4.0 \times 10^4\text{ Bq}\). What is the half-life of the nuclide?
A.1.5 hours
B.2.0 hours
C.3.0 hours
D.4.5 hours
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解題
The ratio of the final activity to the initial activity is \(\frac{4.0 \times 10^4}{3.2 \times 10^5} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly three half-lives have elapsed during the \(6.0\text{ hours}\). Therefore, the half-life \(T_{1/2}\) is given by \(3 \times T_{1/2} = 6.0\text{ hours}\), which yields \(T_{1/2} = 2.0\text{ hours}\).
評分準則
1 mark for identifying that the activity has decreased by 3 half-lives and dividing the total time by 3 to get 2.0 hours.
題目 32 · 選擇題
1 分
A baryon has a net charge of \(+e\). It is composed of three quarks of flavors selected from up (\(u\)), down (\(d\)), and strange (\(s\)). The baryon contains exactly one strange quark. What is the flavor composition of the other two quarks?
A.both are down quarks
B.both are up quarks
C.one is an up quark and one is a down quark
D.one is an up quark and one is a strange quark
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解題
A baryon contains three quarks. Let the charges of the three quarks be \(q_1\), \(q_2\), and \(q_3\). We are given that one of the quarks is a strange quark (\(s\)), which has a charge of \(-\frac{1}{3}e\). The net charge of the baryon is \(+1e\). Therefore, \(q_1 + q_2 - \frac{1}{3}e = +1e\), which simplifies to \(q_1 + q_2 = +\frac{4}{3}e\). The other two quarks must be selected from up (charge \(+\frac{2}{3}e\)) and down (charge \(-\frac{1}{3}e\)) quarks. The only combination of two quarks that sums to \(+\frac{4}{3}e\) is two up quarks (\(+\frac{2}{3}e + \frac{2}{3}e = +\frac{4}{3}e\)). Thus, both other quarks are up quarks (giving a total quark composition of \(uus\)).
評分準則
1 mark for setting up the charge conservation equation and identifying that the remaining two quarks must both be up quarks to achieve a net charge of +e.
題目 33 · 選擇題
1 分
A student wants to determine the density \(\rho\) of a metal cylinder. The mass \(m\) is measured as \(m = (120.0 \pm 0.5)\text{ g}\). The diameter \(d\) is measured as \(d = (2.50 \pm 0.02)\text{ cm}\). The length \(L\) is measured as \(L = (6.00 \pm 0.05)\text{ cm}\). What is the percentage uncertainty in the calculated value of the density?
A.1.7%
B.2.1%
C.2.9%
D.3.7%
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解題
The formula for the density of a cylinder is \(\rho = \frac{4m}{\pi d^2 L}\). The fractional uncertainty in density is given by \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Substituting the values: \(\frac{\Delta m}{m} = \frac{0.5}{120.0} \approx 0.00417\), \(\frac{\Delta d}{d} = \frac{0.02}{2.50} = 0.00800\), and \(\frac{\Delta L}{L} = \frac{0.05}{6.00} \approx 0.00833\). Thus, \(\frac{\Delta \rho}{\rho} = 0.00417 + 2(0.00800) + 0.00833 = 0.0285\). Converting this to a percentage gives \(2.85\%\), which rounds to \(2.9\%\).
評分準則
Award 1 mark for the correct calculation of the percentage uncertainty by summing the relative uncertainties with the correct multiplier for the diameter.
題目 34 · 選擇題
1 分
A picture of weight \(W\) is suspended symmetrically from a hook by a light string of total length \(L\). The two ends of the string are attached to two points on the top edge of the picture frame that are a distance \(d\) apart. The tension in the string is \(T\). Which expression represents the tension \(T\) in terms of \(W\), \(L\), and \(d\)?
A.\(T = \frac{W d}{2 \sqrt{L^2 - d^2}}\)
B.\(T = \frac{W L}{2 \sqrt{L^2 - d^2}}\)
C.\(T = \frac{W L}{\sqrt{L^2 - d^2}}\)
D.\(T = \frac{W \sqrt{L^2 - d^2}}{2 L}\)
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解題
The string is divided into two halves, each of length \(L/2\). The horizontal distance from the center hook to each attachment point is \(d/2\). The vertical height \(h\) of the triangle formed by the string is \(h = \sqrt{(L/2)^2 - (d/2)^2} = \frac{1}{2}\sqrt{L^2 - d^2}\). The sine of the angle \(\theta\) with the horizontal is \(\sin\theta = \frac{h}{L/2} = \frac{\sqrt{L^2 - d^2}}{L}\). For vertical equilibrium of the picture, \(2 T \sin\theta = W\), which gives \(T = \frac{W}{2 \sin\theta} = \frac{W L}{2 \sqrt{L^2 - d^2}}\).
評分準則
Award 1 mark for identifying the correct trigonometric ratio from the geometry of the string and resolving forces vertically to obtain the correct expression.
題目 35 · 選擇題
1 分
An electric pump raises water through a vertical height of \(12\text{ m}\) and discharges it with a speed of \(8.0\text{ m s}^{-1}\) through a pipe. The pump has an efficiency of \(65\%\). If the pump consumes electrical power at a rate of \(3.0\text{ kW}\), what is the mass of water discharged per second? (Take \(g = 9.81\text{ m s}^{-2}\))
A.\(8.3\text{ kg s}^{-1}\)
B.\(13\text{ kg s}^{-1}\)
C.\(17\text{ kg s}^{-1}\)
D.\(20\text{ kg s}^{-1}\)
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解題
The useful power output of the pump is \(P_{\text{out}} = 0.65 \times 3000\text{ W} = 1950\text{ W}\). This power is used to increase both the gravitational potential energy and kinetic energy of the water per second: \(P_{\text{out}} = \frac{m}{t} \left( g h + \frac{1}{2} v^2 \right)\). Substituting the values: \(1950 = \frac{m}{t} \left( 9.81 \times 12 + \frac{1}{2} \times 8.0^2 \right) = \frac{m}{t} (117.72 + 32.0) = \frac{m}{t} (149.72)\). Solving for mass per second gives \(\frac{m}{t} = \frac{1950}{149.72} \approx 13\text{ kg s}^{-1}\).
評分準則
Award 1 mark for calculating the useful output power and correctly setting up and solving the conservation of energy equation per unit time.
題目 36 · 選擇題
1 分
Two wires, X and Y, are made of the same material. Wire X has a length \(L\) and a diameter \(d\). Wire Y has a length \(2L\) and a diameter \(2d\). Both wires are suspended vertically and support loads. Wire X supports a load of weight \(F\), and wire Y supports a load of weight \(3F\). What is the ratio of the strain in wire X to the strain in wire Y?
A.\(\frac{2}{3}\)
B.\(\frac{3}{4}\)
C.\(\frac{4}{3}\)
D.\(\frac{8}{3}\)
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解題
Since both wires are made of the same material, they have the same Young Modulus \(E\). Because \(\text{strain} = \frac{\text{stress}}{E}\), the ratio of their strains is equal to the ratio of their stresses. Stress \(\sigma = \frac{\text{Force}}{\text{Area}}\). For wire X, \(\sigma_X = \frac{F}{\pi d^2 / 4} = \frac{4F}{\pi d^2}\). For wire Y, \(\sigma_Y = \frac{3F}{\pi (2d)^2 / 4} = \frac{3F}{\pi d^2}\). Therefore, the ratio of the strains is \(\frac{\sigma_X}{\sigma_Y} = \frac{4}{3}\).
評分準則
Award 1 mark for expressing strain in terms of stress and Young Modulus, and calculating the correct ratio of 4/3.
題目 37 · 選擇題
1 分
Two coherent sound sources, \(S_1\) and \(S_2\), emit waves with amplitudes \(2A\) and \(A\) respectively. A microphone is placed at point \(P\) where the path difference between the waves from the two sources is \(1.5\lambda\), where \( \lambda \) is the wavelength. Initially, the two sources emit in phase, and the sound intensity detected at \(P\) is \(I_0\). The phase of source \(S_1\) is then shifted by \(180^\circ\) (\(\pi\) radians) while keeping the amplitudes constant. What is the new intensity detected at \(P\)?
A.\(1.5 I_0\)
B.\(3.0 I_0\)
C.\(4.5 I_0\)
D.\(9.0 I_0\)
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解題
Initially, the path difference of \(1.5\lambda\) causes a phase difference of \(3\pi\) radians, resulting in destructive interference. The initial resultant amplitude is \(2A - A = A\), meaning the initial intensity is \(I_0 = k A^2\). When source \(S_1\) is phase-shifted by \(180^\circ\), the waves now arrive in phase at \(P\), resulting in constructive interference. The new resultant amplitude is \(2A + A = 3A\). The new intensity is therefore \(I = k (3A)^2 = 9 k A^2 = 9.0 I_0\).
評分準則
Award 1 mark for identifying the shift from destructive to constructive interference and using the relationship between intensity and amplitude squared to find \(9.0 I_0\).
題目 38 · 選擇題
1 分
A neutral \(\Sigma^0\) baryon has a quark composition of \(\text{uds}\) (up, down, strange). It decays into a neutral \(\Lambda^0\) baryon (also with quark composition \(\text{uds}\)) and a photon \(\gamma\) via the decay \(\Sigma^0 \to \Lambda^0 + \gamma\). Which row correctly describes the interaction responsible for this decay and the charge of the \(\Sigma^0\) baryon in terms of the elementary charge \(e\)?
A.Strong interaction, charge 0
B.Weak interaction, charge +1
C.Weak interaction, charge 0
D.Electromagnetic interaction, charge 0
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解題
The decay produces a photon \(\gamma\) without changing the quark flavors, which means it is mediated by the electromagnetic interaction. The quark composition of \(\Sigma^0\) is \(\text{uds}\). Given the charges of the quarks are \(u = +2/3 e\), \(d = -1/3 e\), and \(s = -1/3 e\), the total charge is \(+2/3 e - 1/3 e - 1/3 e = 0\).
評分準則
Award 1 mark for identifying that the emission of a photon represents an electromagnetic interaction and that the total quark charge of the \(\Sigma^0\) baryon is 0.
題目 39 · 選擇題
1 分
A ball is thrown vertically upwards from the edge of a cliff with an initial speed \(u\). It rises to a maximum height and then falls, passing the edge of the cliff and continuing to the ground at the base of the cliff. The total time of flight is \(T\), and the ground is at a distance \(H\) vertically below the edge of the cliff. Assuming air resistance is negligible and taking the downwards direction as positive, which equation correctly relates \(H\), \(u\), and \(T\)?
A.\(H = u T + \frac{1}{2} g T^2\)
B.\(H = -u T + \frac{1}{2} g T^2\)
C.\(H = u T - \frac{1}{2} g T^2\)
D.\(H = -u T - \frac{1}{2} g T^2\)
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解題
Using the kinematic equation for displacement \(s = u_0 T + \frac{1}{2} a T^2\). Setting the edge of the cliff as the origin and downwards as positive: the displacement to the ground is \(s = +H\) because the ground is below the starting point. The initial velocity is directed upwards, so \(u_0 = -u\). The acceleration is downwards due to gravity, so \(a = +g\). Substituting these values yields \(H = -u T + \frac{1}{2} g T^2\).
評分準則
Award 1 mark for applying sign conventions correctly to the displacement, initial velocity, and acceleration to obtain the correct expression.
題目 40 · 選擇題
1 分
A uniform metal wire of resistance \(R\) is stretched until its length is increased by \(25\%\). Assuming that the density and resistivity of the metal remain unchanged during stretching, what is the new resistance of the wire?
A.\(1.25 R\)
B.\(1.50 R\)
C.\(1.56 R\)
D.\(1.80 R\)
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解題
Let the initial length be \(L\) and area be \(A\). The new length is \(L' = 1.25 L\). Since the volume of the wire remains constant during stretching, \(A L = A' L' \implies A' = A \frac{L}{1.25 L} = 0.8 A\). The new resistance is given by \(R' = \rho \frac{L'}{A'} = \rho \frac{1.25 L}{0.8 A} = \frac{1.25}{0.8} R = 1.5625 R \approx 1.56 R\).
評分準則
Award 1 mark for applying conservation of volume to determine the new cross-sectional area and calculating the correct final resistance.
卷二 (AS Level 結構題)
Answer all 6 structured questions. Show all working and use appropriate units where necessary.
6 題目 · 60 分
題目 1 · structured
10 分
An experiment is performed to determine the resistivity \(\rho\) of a uniform metal wire of length \(L\), resistance \(R\), and diameter \(d\). The resistivity is given by the formula:
\(\rho = \frac{\pi R d^2}{4L\)}
The measurements obtained and their absolute uncertainties are: \(R = (4.50 \pm 0.05)\ \Omega\) \(d = (0.38 \pm 0.01)\text{ mm\)} \(L = (1.250 \pm 0.002)\text{ m\)}
(a) Distinguish between systematic errors and random errors, and state how the effect of random errors can be reduced. [3]
(b) (i) Calculate the value of \(\rho\) in \(\Omega\text{ m}\) to a suitable number of significant figures. [3] (ii) Calculate the percentage uncertainty in \(\rho\). [3] (iii) Determine the absolute uncertainty in \(\rho\). [1]
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解題
(a) - Systematic errors cause measurements to deviate from the true value by a constant amount in the same direction each time, whereas random errors cause measurements to vary unpredictably around the true value. - The effect of random errors can be reduced by taking multiple readings and calculating a mean value.
(b) (i) Convert diameter: \(d = 0.38 \times 10^{-3}\text{ m\)} \(\rho = \frac{\pi \times 4.50 \times (0.38 \times 10^{-3})^2}{4 \times 1.250\)} \(\rho = \frac{\pi \times 4.50 \times 1.444 \times 10^{-7}}{5.000} = 4.083 \times 10^{-7}\ \Omega\text{ m\)} Since \(d\) is measured to 2 significant figures, the final resistivity should be quoted to 2 significant figures: \(\rho = 4.1 \times 10^{-7}\ \Omega\text{ m\)}
(a) - [1] Systematic error defined (constant deviation in one direction). - [1] Random error defined (unpredictable fluctuations above and below the true value). - [1] State that random errors are reduced by taking repeat measurements and finding the average.
(b)(i) - [1] Correct substitution of numbers into the formula with correct power of ten for \(d\). - [1] Calculation of \(4.08 \times 10^{-7}\) or \(4.1 \times 10^{-7}\). - [1] Answer given to 2 s.f. with correct units (\(\Omega\text{ m}\)) consistent with the least precise input data (\(d\)).
(b)(ii) - [1] Summing fractional uncertainties with factor of 2 for diameter (\(\frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)). - [1] Correctly calculating individual percentage uncertainties (1.11%, 5.26%, 0.16%). - [1] Total percentage uncertainty calculated as 6.5% (allow 6.5% to 6.6%).
(b)(iii) - [1] Correct absolute uncertainty calculated and rounded to 1 s.f. (\(0.3 \times 10^{-7}\ \Omega\text{ m}\) or 2 s.f. \(0.27 \times 10^{-7}\ \Omega\text{ m}\)).
題目 2 · structured
10 分
A ball is projected from the edge of a cliff at an angle of \(35.0^\circ\) above the horizontal with a speed of \(15.0\text{ m s}^{-1}\). The cliff is at a height of \(25.0\text{ m}\) above the sea level. Air resistance is negligible.
(a) Show that the vertical component of the initial velocity is \(8.60\text{ m s}^{-1}\) and the horizontal component is \(12.3\text{ m s}^{-1}\). [2]
(b) Calculate the maximum height reached by the ball above the launch point. [2]
(c) Determine the total time taken for the ball to reach the sea. [3]
(d) Calculate the speed of the ball immediately before it hits the sea. [3]
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解題
(a) Vertical component: \(u_y = u \sin\theta = 15.0 \sin(35.0^\circ) = 8.6036 \approx 8.60\text{ m s}^{-1}\) Horizontal component: \(u_x = u \cos\theta = 15.0 \cos(35.0^\circ) = 12.287 \approx 12.3\text{ m s}^{-1}\)
(b) At maximum height, the vertical velocity \(v_y = 0\). \(v_y^2 = u_y^2 - 2g y_{max}\) \(0 = (8.604)^2 - 2(9.81)y_{max}\) \(y_{max} = \frac{74.03}{19.62} = 3.77\text{ m}\)
(c) The total vertical displacement when the ball hits the sea is \(y = -25.0\text{ m}\) (taking upwards as positive). \(y = u_y t - \frac{1}{2}g t^2\) \(-25.0 = 8.604 t - 4.905 t^2\) \(4.905 t^2 - 8.604 t - 25.0 = 0\) Using the quadratic formula: \(t = \frac{8.604 \pm \sqrt{(-8.604)^2 - 4(4.905)(-25.0)}}{2(4.905)}\) \(t = \frac{8.604 \pm \sqrt{74.03 + 490.5}}{9.81}\) \(t = \frac{8.604 + 23.76}{9.81} = 3.30\text{ s}\) (discarding negative root)
(d) Using conservation of energy: \(\text{Initial total energy per unit mass} = \frac{1}{2} u^2 + gh\) \(E_k + E_p = \frac{1}{2}(15.0)^2 + (9.81 \times 25.0) = 112.5 + 245.25 = 357.75\text{ J kg}^{-1}\) At sea level, \(E_p = 0\): \(\frac{1}{2} v^2 = 357.75\) \(v = \sqrt{2 \times 357.75} = \sqrt{715.5} = 26.8\text{ m s}^{-1}\)
Alternatively, calculate the components of final velocity: \(v_x = 12.29\text{ m s}^{-1}\) \(v_y^2 = u_y^2 - 2gy = (8.604)^2 - 2(9.81)(-25.0) = 74.03 + 490.5 = 564.53\text{ m}^2\text{ s}^{-2}\) \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12.29^2 + 564.53} = \sqrt{151.04 + 564.53} = \sqrt{715.57} = 26.8\text{ m s}^{-1}\)
評分準則
(a) - [1] Vertical component shown correctly using \(15.0 \sin(35.0^\circ)\). - [1] Horizontal component shown correctly using \(15.0 \cos(35.0^\circ)\).
(b) - [1] Use of \(v^2 = u^2 + 2as\) with vertical component and \(v_y = 0\). - [1] Correct calculation of height as \(3.77\text{ m}\) (accept \(3.8\text{ m}\)).
(c) - [1] Use of \(s = ut + \frac{1}{2}at^2\) with correct vertical displacement value of \(-25.0\text{ m}\) (or breaking the motion into two parts). - [1] Setting up a valid quadratic equation or solving for time to reach max height plus time to fall from max height. - [1] Correct calculation of time as \(3.30\text{ s}\) (accept \(3.3\text{ s}\)).
(d) - [1] Stating a correct energy conservation statement or kinematics equation for vertical velocity component. - [1] Correct determination of the vertical component of final velocity (\(v_y = 23.8\text{ m s}^{-1}\)) OR direct energy method setup. - [1] Correct speed calculation of \(26.8\text{ m s}^{-1}\) (allow \(26.7 - 26.9\text{ m s}^{-1}\)).
題目 3 · structured
10 分
A small uniform sphere of mass \(m = 0.85\text{ kg}\) is suspended from a ceiling by a light inextensible string. A horizontal force \(F\) is applied to the sphere so that the string makes an angle of \(28^\circ\) with the vertical. The system is in static equilibrium.
(a) State the two conditions required for an object to be in static equilibrium. [2]
(b) Describe and sketch (or describe the direction of) the three forces acting on the sphere, identifying each by name. [2]
(c) Calculate: (i) the tension \(T\) in the string. [3] (ii) the magnitude of the horizontal force \(F\). [2]
(d) The force \(F\) is adjusted so that the angle between the string and the vertical increases. State and explain the effect of this change on the tension \(T\). [1]
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解題
(a) 1. The resultant force acting on the object must be zero in any direction (translational equilibrium). 2. The resultant torque/moment about any axis of rotation must be zero (rotational equilibrium).
(b) The three forces are: 1. Weight (\(W = mg\)), acting vertically downwards from the center of mass of the sphere. 2. Tension (\(T\)), acting along the line of the string, directed upwards and away from the sphere. 3. Applied force (\(F\)), acting horizontally to the side.
(d) Since \(T = \frac{W}{\cos\theta}\), as the angle \(\theta\) increases, \\cos\theta\) decreases. Therefore, for a constant weight \(W\), the tension \(T\) in the string must increase.
評分準則
(a) - [1] State that the resultant/net force in any direction is zero. - [1] State that the net torque/moment about any point is zero.
(b) - [1] Identify all three forces: Weight downwards, Tension along string (up/away), and horizontal applied force. - [1] Show or describe their directions correctly meeting at the center of the sphere.
(c)(i) - [1] Correct calculation of the weight of the sphere (\(W = 8.34\text{ N}\) or \(8.3\text{ N}\)). - [1] Resolving forces vertically to get \(T \cos(28^\circ) = mg\). - [1] Correct calculation of tension \(T = 9.44\text{ N}\) (accept \(9.4\text{ N}\)).
(c)(ii) - [1] Resolving forces horizontally to get \(F = T \sin(28^\circ)\) or \(F = W \tan(28^\circ)\). - [1] Correct calculation of \(F = 4.43\text{ N}\) (accept \(4.4\text{ N}\)).
(d) - [1] Correctly states tension increases and explains that \(\cos\theta\) decreases as \(\theta\) increases, and since \(T = W/\cos\theta\), \(T\) must increase.
題目 4 · structured
10 分
A roller-coaster car of mass \(450\text{ kg}\) starts from rest at point A, which is at a height of \(35.0\text{ m}\) above the ground. It travels down a curved track to point B at ground level, and then travels up a second hill to point C, which is at a height of \(22.0\text{ m}\) above the ground.
(a) Calculate the decrease in gravitational potential energy of the car as it travels from A to B. [2]
(b) Assuming the track is frictionless between A and B, calculate the speed of the car at B. [2]
(c) In reality, a constant average resistive force of \(380\text{ N}\) acts on the car along the track from A to B. The length of the track between A and B is \(60.0\text{ m}\). (i) Calculate the work done against the resistive force. [2] (ii) Calculate the actual speed of the car when it reaches B. [2]
(d) At point C, the car's actual speed is measured to be \(8.50\text{ m s}^{-1}\). Calculate the total thermal energy generated by resistive forces during the entire journey from A to C. [2]
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解題
(a) \(\Delta E_p = m g \Delta h = 450 \times 9.81 \times 35.0 = 154,507.5\text{ J} \approx 1.55 \times 10^5\text{ J\)}
(b) All GPE is converted to KE: \(\frac{1}{2} m v^2 = \Delta E_p\) \(\frac{1}{2} \times 450 \times v^2 = 154,507.5\) \(v = \sqrt{2 \times 9.81 \times 35.0} = 26.2\text{ m s}^{-1\)}
(c) (i) \(W_{\text{resistive}} = F d = 380 \times 60.0 = 22,800\text{ J} = 2.28 \times 10^4\text{ J\)}
(ii) Actual kinetic energy at B: \(E_{k\text{, actual}} = \Delta E_p - W_{\text{resistive}} = 154,507.5 - 22,800 = 131,707.5\text{ J\)} \(\frac{1}{2} m v_{\text{actual}}^2 = 131,707.5\) \(v_{\text{actual}} = \sqrt{\frac{2 \times 131,707.5}{450}} = 24.2\text{ m s}^{-1\)}
(d) Total energy at the start (at A): \(E_A = m g h_A = 450 \times 9.81 \times 35.0 = 154,507.5\text{ J\)}
Total energy at C: \(E_C = m g h_C + \frac{1}{2} m v_C^2 = (450 \times 9.81 \times 22.0) + \left(\frac{1}{2} \times 450 \times 8.50^2\right)\) \(E_C = 97,119 + 16,256.25 = 113,375.25\text{ J\)}
Thermal energy generated = loss in mechanical energy: \(Q = E_A - E_C = 154,507.5 - 113,375.25 = 41,132.25\text{ J} \approx 4.11 \times 10^4\text{ J\)} (or \(41.1\text{ kJ}\))
評分準則
(a) - [1] Use of \(\Delta E_p = mg\Delta h\). - [1] Correct calculation of \(1.55 \times 10^5\text{ J}\) (or \(1.5 \times 10^5\text{ J}\)).
(b) - [1] Equating \(E_k\) to \(E_p\) (\(\frac{1}{2}mv^2 = mgh\)). - [1] Correct calculation of speed as \(26.2\text{ m s}^{-1}\).
(c)(i) - [1] Use of \(W = Fd\). - [1] Correct calculation of \(22,800\text{ J}\) (or \(2.3 \times 10^4\text{ J}\)).
(c)(ii) - [1] Subtracting the work done against friction from initial GPE to find the final kinetic energy. - [1] Correct calculation of actual speed as \(24.2\text{ m s}^{-1}\) (allow \(24.0 - 24.5\text{ m s}^{-1}\)).
(d) - [1] Stating and substituting into the conservation of energy equation: \(E_A = E_C + E_{\text{thermal}}\). - [1] Correct calculation of thermal energy as \(4.11 \times 10^4\text{ J}\) (accept \(4.1 \times 10^4\text{ J}\) or \(41\text{ kJ}\)).
題目 5 · structured
10 分
A steel wire of length \(2.20\text{ m}\) and cross-sectional area \(1.50 \times 10^{-6}\text{ m}^2\) is suspended vertically. A load of \(180\text{ N}\) is applied to its lower end. The Young modulus of steel is \(2.00 \times 10^{11}\text{ Pa}\).
(a) Define *Young modulus*. [1]
(b) Show that the extension of the wire under this load is approximately \(1.3\text{ mm}\). [3]
(c) Calculate the elastic strain energy stored in the wire when carrying this load, assuming Hooke's law is obeyed. [2]
(d) The load is increased until the wire behaves plastically. (i) State the difference between *elastic deformation* and *plastic deformation*. [2] (ii) Describe the feature of a force-extension graph that shows the permanent deformation of the wire when it is loaded into the plastic region and then unloaded. [2]
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解題
(a) Young modulus is defined as the ratio of tensile stress to tensile strain: \(E = \frac{\text{Stress}}{\text{Strain}}\) (provided the proportional limit is not exceeded).
(c) Strain energy stored is: \(E_s = \frac{1}{2} F x = \frac{1}{2} \times 180 \times 1.32 \times 10^{-3} = 0.119\text{ J}\) (or \(0.12\text{ J}\))
(d) (i) - Elastic deformation: the material returns to its original length/shape when the load/deforming force is removed. - Plastic deformation: the material does not return to its original length/shape when the load is removed; there is permanent extension/deformation.
(ii) - When loaded past the elastic limit, the line on the force-extension graph curves and its gradient decreases. - When unloaded, the unloading curve is a straight line parallel to the initial linear portion of the loading curve, which intercepts the extension axis at a value greater than zero, showing a permanent extension.
(b) - [1] Use of \(E = \frac{FL}{Ax}\) or equivalent step-by-step stress and strain calculations. - [1] Correct substitution of all given quantities with correct powers of 10. - [1] Correctly calculates \(1.32 \times 10^{-3}\text{ m}\) and rounds/states it to be \(1.3\text{ mm}\).
(c) - [1] Use of \(E_s = \frac{1}{2}Fx\) (or \(E_s = \frac{1}{2}kx^2\)). - [1] Correct calculation of \(0.119\text{ J}\) or \(0.12\text{ J}\).
(d)(i) - [1] Clarifies that elastic deformation means returning to original shape when force is removed. - [1] Clarifies that plastic deformation means permanent change in shape/does not return to original shape.
(d)(ii) - [1] Mentions that the unloading line is parallel to the straight-line section of the loading curve. - [1] Mentions that the unloading line does not return to the origin, showing a non-zero permanent extension.
題目 6 · structured
10 分
A glass tube, closed at one end and open at the other, has a small loudspeaker placed near the open end. The frequency of the sound emitted by the loudspeaker is slowly increased from zero. The first resonance is observed at a frequency of \(280\text{ Hz}\). The speed of sound in air is \(340\text{ m s}^{-1}\).
(a) Explain how a stationary wave is formed in the tube. [3]
(b) (i) Draw the stationary wave pattern (displacement) inside the tube for this first resonance. Label any displacement nodes with 'N' and antinodes with 'A'. [2] (ii) Calculate the length \(L\) of the tube, assuming the antinode is exactly at the open end. [2]
(c) Determine the next frequency at which a resonance will be observed in this tube. [3]
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解題
(a) - Sound waves travel down the tube from the loudspeaker and are reflected at the closed end. - The incident and reflected waves, which have the same frequency and speed, travel in opposite directions and meet/superpose. - This produces a stationary wave pattern with points of zero displacement (nodes) and maximum displacement (antinodes).
(b) (i) - The sketch should show a node (N) at the closed end and an antinode (A) at the open end. - The pattern represents a quarter of a wavelength (a single loop tapering down to the closed end).
(ii) For the fundamental mode (first resonance): \(L = \frac{\lambda}{4\)} Since \(v = f \lambda \implies \lambda = \frac{v}{f} = \frac{340}{280} = 1.214\text{ m\)} \(L = \frac{1.214}{4} = 0.304\text{ m\)} (or \(30.4\text{ cm\) or \(0.30\text{ m\) to 2 s.f.)
(c) For a tube closed at one end, the possible resonant modes are odd harmonics (\(f, 3f, 5f, \dots\)). The next resonance is the 3rd harmonic (first overtone): \(f_{\text{next}} = 3 \times f_{\text{fundamental}} = 3 \times 280 = 840\text{ Hz\)}
Alternatively, for the next mode, \(L = \frac{3\lambda'}{4} \implies \lambda' = \frac{4L}{3}\): \(f_{\text{next}} = \frac{v}{\lambda'} = \frac{3v}{4L} = \frac{3 \times 340}{4 \times 0.3036} = 840\text{ Hz\)}
評分準則
(a) - [1] Sound waves travel down the tube and reflect at the closed end. - [1] Wave traveling down the tube and reflected wave superpose/interfere. - [1] The two waves must have the same frequency/amplitude and travel in opposite directions.
(b)(i) - [1] Correct drawing of a quarter-wavelength pattern (one node and one antinode). - [1] Node correctly labeled 'N' at the closed end and antinode 'A' at the open end.
(b)(ii) - [1] Correct link between wavelength and length of tube: \(L = \lambda / 4\). - [1] Correct calculation of length \(L = 0.304\text{ m}\) (accept \(0.30\text{ m}\) to \(0.31\text{ m}\)).
(c) - [1] Identifies that the next mode of resonance corresponds to a three-quarter wavelength in the tube (\(L = 3\lambda/4\) or 3rd harmonic). - [1] Relates the next frequency to the fundamental frequency: \(f' = 3f\). - [1] Correctly calculates \(f' = 840\text{ Hz}\).
Paper 3 (Advanced Practical Skills)
Complete two practical experimental questions, including recording raw measurements, plotting graphs, and analyzing limitations.
2 題目 · 40 分
題目 1 · practical
20 分
Apparatus List:
Metre rule (uniform, to be pivoted)
Elastic band (unstretched length approximately \(8\text{ cm}\) to \(10\text{ cm}\))
Two retort stands with bosses and clamps
A G-clamp or heavy weight to secure the stands to the bench
A loop of thread to act as a pivot at the \(10.0\text{ cm}\) mark of the metre rule
A horizontal rod (e.g. a clamp arm or a metal rod) clamped to the first stand to support the thread loop
A mass hanger with slotted masses to make a total load of \(200\text{ g}\)
A second metre rule or a half-metre rule to measure heights
In this experiment, you will investigate the equilibrium of a pivoted wooden rule supported by an inclined elastic band.
Procedure:
Measure and record the unstretched length \(L_0\) of the elastic band.
Set up the apparatus as described below:
Place the horizontal rod in a clamp on the first stand.
Loop the thread around the \(10.0\text{ cm}\) mark of the metre rule and suspend it from the horizontal rod. This acts as a pivot.
Attach one end of the elastic band to the \(90.0\text{ cm}\) mark of the rule.
Attach the other end of the elastic band to a clamp on the second stand.
Adjust the heights of the clamps so that the metre rule is horizontal.
Measure and record the height \(h_0\) of the \(90.0\text{ cm}\) mark above the bench.
Suspend the \(200\text{ g}\) mass at a distance \(x = 20.0\text{ cm}\) from the pivot (the \(10.0\text{ cm}\) mark).
Measure and record the new height \(h\) of the \(90.0\text{ cm}\) mark above the bench.
Calculate the deflection \(y = h_0 - h\).
Vary \(x\) in the range \(15.0\text{ cm} \le x \le 80.0\text{ cm}\) and measure \(h\). Record your results in a table. Include columns for \(x\), \(h\), and \(y = h_0 - h\).
Plot a graph of \(y\) on the y-axis against \(x\) on the x-axis.
Determine the gradient and y-intercept of the line.
The quantities \(y\) and \(x\) are related by the equation: \(y = P x + Q\) where \(P\) and \(Q\) are constants. Use your answers from the previous step to determine the values of \(P\) and \(Q\). Include appropriate units.
Estimate the absolute uncertainty in your measurement of \(h\), justifying your choice.
Calculate the percentage uncertainty in your smallest value of \(h\).
(i) Absolute uncertainty in \(h\) is estimated as \(\pm 0.2\text{ cm}\) because of the difficulty in ensuring the vertical alignment of the rule and parallax errors in reading the scale.
(a) [1 mark] Unstretched length \(L_0\) recorded to nearest mm with correct unit.
(b) [1 mark] Initial height \(h_0\) recorded to nearest mm with correct unit.
(c) [1 mark] First measurement of \(h\) recorded to nearest mm and correct deflection \(y = h_0 - h\) calculated.
(d) Table of results (6 marks):
Range and number of readings: At least 6 sets of readings of \(x\) and \(h\) taken over a range of at least \(50.0\text{ cm}\). [1]
Column headings: Heading and unit for each column must be present in the format "variable / unit" (e.g., \(x / \text{cm}\), \(h / \text{cm}\), \(y / \text{cm}\)). [1]
Consistency: All raw readings of \(x\) and \(h\) must be recorded to the nearest mm (e.g., \(15.0\text{ cm}\), not \(15\text{ cm}\)). [1]
Significant figures: Significant figures/decimal places in \(y\) must be consistent with the precision of raw \(h\) and \(h_0\). [1]
Calculations: All calculations of \(y\) must be mathematically correct. [1]
Quality of data: All plotted points must lie close to a straight line. [1]
(e) Graph (4 marks):
Axes: Linear scales with sensible intervals (not multiples of 3, 7, etc.). Labeled with units. [1]
Plotting: All points plotted to within half a small square on the grid. [1]
Line of best fit: Single sharp line drawn with balanced distribution of points above and below the line. [1]
Quality: No points scatter more than \(1.0\text{ cm}\) (on the graph scale) from the best-fit line. [1]
(f) Gradient and Intercept (2 marks):
Gradient: Correct read-offs from a triangle with hypotenuse length at least half the length of the line. [1]
y-intercept: Correctly calculated from \(y = m x + c\) or read directly from the y-axis if \(x=0\). [1]
(g) Constants (2 marks):
Values of \(P\) and \(Q\) stated to 2 or 3 significant figures. [1]
Correct units given: \(P\) is dimensionless (or \(\text{cm cm}^{-1}\)), and \(Q\) has units of length (e.g., \(\text{cm}\) or \(\text{m}\)). [1]
(h) Uncertainty (2 marks):
(i) Absolute uncertainty in \(h\) estimated as \(\pm 0.1\text{ cm}\) to \(\pm 0.3\text{ cm}\) with a valid reason (e.g., parallax/scale alignment). [1]
(ii) Percentage uncertainty calculated correctly to 2 s.f. using \(\frac{\Delta h}{h_{\text{smallest}}} \times 100\%\). [1]
In this experiment, you will investigate the relationship between the length of the suspending threads and the period of oscillation of a horizontally suspended metre rule.
Procedure:
Set up the apparatus as described below:
Suspend the metre rule horizontally using two vertical threads of equal length \(L\).
Attach the threads to the metre rule at the \(10.0\text{ cm}\) mark and the \(90.0\text{ cm}\) mark using small pieces of tape.
Clamp the upper ends of the threads so that they are parallel and vertical.
Adjust the thread length \(L\) to approximately \(30.0\text{ cm}\).
Measure and record the length \(L_1\) of one of the suspending threads to the nearest mm.
Displace the metre rule slightly in a direction perpendicular to its length, and release it so that it oscillates in a horizontal plane.
Place a fiducial marker on the bench directly below the \(50.0\text{ cm}\) mark of the rule at its equilibrium position.
Measure and record the time \(t_1\) for at least 10 complete oscillations.
Repeat the measurement and record the average time \(t_{\text{avg1}}\).
Calculate the period \(T_1\) of the oscillations.
Estimate the percentage uncertainty in your measurement of \(t_1\), showing your working and justifying your choice of absolute uncertainty.
Adjust the clamps to increase the length \(L\) of both threads to approximately \(60.0\text{ cm}\).
Measure and record the new thread length \(L_2\).
Measure and record the time \(t_2\) for at least 10 complete oscillations (repeated).
Calculate the new period \(T_2\).
It is suggested that the relationship between \(T\) and \(L\) is: \[T^2 = k L\] where \(k\) is a constant.
Calculate the two values of \(k\).
Explain whether your results support the suggested relationship. Justify your conclusion by comparing the percentage difference between your two values of \(k\) with your estimated percentage uncertainty.
Identify and describe four sources of uncertainty or systematic/random errors in this experiment, and suggest four improvements that could be made to reduce these uncertainties/limitations. Present your answers in a table.
Since \(0.74\% < 1.8\%\) (the estimated percentage uncertainty), the results support the suggested relationship.
Table of Limitations and Improvements:
Limitations / Sources of UncertaintySuggested ImprovementsTwo values of \(L\) are insufficient to draw a firm conclusion.Take multiple values of \(L\) and plot a graph of \(T^2\) against \(L\).The metre rule oscillates with a slight twisting/torsional motion.Attach threads to the rule in a way that constrains motion to one vertical plane, or use a wider attachment loop.Difficulty in identifying the exact start/end of an oscillation by eye.Use a light gate and a data logger to record the period of oscillation.The threads might not be perfectly vertical and parallel.Use a set square to verify verticality and measure the distance between threads at the top and bottom.
評分準則
(a)(i) [1 mark] First length \(L_1\) recorded to nearest mm with correct unit.
(a)(ii) [2 marks] Raw time \(t_1\) recorded to at least 0.01 s with repeats. Award 1 mark if repeats are missing.
(a)(iii) [1 mark] Period \(T_1\) calculated correctly.
(b) [1 mark] Absolute uncertainty in \(t_1\) chosen within \(0.1\text{ s} \le \Delta t \le 0.5\text{ s}\) with human reaction time mentioned as a justification, and percentage uncertainty calculated correctly.
(c)(i) [1 mark] Second length \(L_2\) recorded to nearest mm.
(c)(ii) [2 marks] Raw time \(t_2\) recorded to at least 0.01 s with repeats.
(c)(iii) [1 mark] Period \(T_2\) calculated correctly.
(d)(i) [1 mark] Correct calculation of both values of \(k\) with units of \(\text{s}^2\text{ m}^{-1}\) (or \(\text{s}^2\text{ cm}^{-1}\)).
(d)(ii) [2 marks] Quantitative comparison between percentage difference and estimated percentage uncertainty. Concludes whether the relationship is supported (supported if percentage difference is less than estimated percentage uncertainty, or less than a stated limit of 10%).
(e) & (f) Limitations and Improvements (8 marks): Award 1 mark for each unique, valid limitation (max 4) and 1 mark for each corresponding valid improvement (max 4). Do not accept generic answers like "human error" or "do it more carefully".
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